MathematicsHard53×since 2002Q3167If z=12−2iz=\frac{1}{2}-2 iz=21−2i is such that ∣z+1∣=αz+β(1+i),i=−1|z+1|=\alpha z+\beta(1+i), i=\sqrt{-1}∣z+1∣=αz+β(1+i),i=−1 and α,β∈R\alpha, \beta \in \mathbb{R}α,β∈R, then α+β\alpha+\betaα+β is equal toA2B−-−4C3D−-−1Check answerSkip